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1. 0.500g of a solid acid was dissolved in water and titrated to endpoint with 3

ID: 794398 • Letter: 1

Question

1.       0.500g of a solid acid was dissolved in water and titrated to endpoint with 30.50 mL of a 0.10M NaOH solution.  What is the equivalent weight of the acid?

A.                   164

B.                   153

C.                   0.161


2.       What is the Normality of 20.00mL of acid solution that is titrated to endpoint with 21.33 mL of a 0.120N base solution

A.                   0.113

B.                   0.120

C.                   0.128


3.       When a base is added to a  buffer composed of HNO2 and NaNO2 , the OH- is neutralized by:

A.                   HNO2

B.                   water

C.                   NO2-


4.       At a temperature other than 25oC, Ni(OH)2 solubility is 2.8 x 10-6. Calculate it's Ksp at this temperature

A.                   2.2 x 10-17

B.                   1.6 x 10-14

C.                   8.8 x 10-17


5.       The combination of (CH3)2NH and (CH3)2NH2Cl can be used to prepare a buffer

True

False


6.       How many grams of a chloride salt would you need to prepare 500 ml of a buffer at pH = 10.62, using 0.100M solution

A.            3.35

B.             0

C.             6.70


Explanation / Answer

1.       0.500g of a solid acid was dissolved in water and titrated to endpoint with 30.50 mL of a 0.10M NaOH solution.  What is the equivalent weight of the acid?

mmol of base = MV = 0.1*30.5= 3.05 mmol of base = 3.05 *10^-3 mol

MW equivalent = mass/equiv = 0.5/(3.05 *10^-3) = 163.934 g/mol

2.       What is the Normality of 20.00mL of acid solution that is titrated to endpoint with 21.33 mL of a 0.120N base solution

N = eq*Mol

N1*V1 = N2*V2

N1*20 = 0.120 * 21.33

N1 = 0.120 * 21.33/20

N1 = 0.12798 Normal

3.       When a base is added to a  buffer composed of HNO2 and NaNO2 , the OH- is neutralized by:

OH- is neutralized by HNO2

since

HNO2 + OH- forms = H2O + NO2-

4.       At a temperature other than 25oC, Ni(OH)2 solubility is 2.8 x 10-6. Calculate it's Ksp at this temperature

Ksp = [Ni2+][OH-]^2

[Ni2+´] = 2.8*10^-6

[OH-] = 2*[Ni(OH)2] = 2*( 2.8*10^-6) = 0.0000056

substitute

Ksp = (2.8*10^-6)(0.0000056^2) = 8.780*10^-17

5.       The combination of (CH3)2NH and (CH3)2NH2Cl can be used to prepare a buffer

(CH3)2-NH and (CH3)-NH2+

both, conjugate acid and weak base are present

so this must be true

Q6 requires pKa data